PHP/SQL错误在输出数据

Question

的PHP：

<?php 
     $pID = filter_input(INPUT_GET, 'pID', FILTER_SANITIZE_NUMBER_INT); 
     if(!$pID) { 
      echo "No pID specified."; 
      exit; 
     } 
     $username = "###"; 
     $password = "####"; 
     $pdo = new PDO('mysql:host=localhost;dbname=###', $username, $password); 
     $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
     $sth = $pdo->prepare(' 
      SELECT D.name, P.lname, P.fname, P.picpath, P.email, C.info, C.date 
       FROM Department D, Professor P, Comment C 
      WHERE D.dID = P.dID 
      AND P.pID = C.pID; 
     '); 
     $sth->execute(array(
      $pID 
     )); 
    ?> 

在同一页面中使用：

<?php 
       if($sth->rowCount() > 0) { 
        $row = $sth->fetch(PDO::FETCH_ASSOC); 
        echo "<img style='float:left;' src='img/profViewIcon.png' alt='Professor {$row['fname']} {$row['lname']} at Drexel' /><h1> <span title='{$row['fname']} {$row['lname']} at Drexel University'> {$row['fname']} {$row['lname']} </h1>"; 
        echo " 
        <div class='professor_pic'> 
         <img src='{$row['picpath']}' /> 
        </div><!-- /professor_pic --> 
        <div class='professor_desc'> 
         <span class='one' style='float:left; padding:5px 0 0 5px;'><strong>Department:</strong> {$row['name']} </span><br> 
         <span class='two' style='float:left; padding:5px 0 0 5px;'><strong>Email:</strong> {$row['email']} </span><br> 
         <span class='two' style='float:right;'> 
          <strong>Rating:</strong><img src='img/fullStar.png'><img src='img/fullStar.png'><img src='img/fullStar.png'><img src='img/halfStar.png'> 
         </span><br> 
         <span class='two' style='float:left; padding:20px 5px 0 5px; font-size:1.50em;'> 
          <img src='img/commentOnProf.png' class='left' />&nbsp;<a href='#addComment' style='font-weight:bold;'>Write A Comment</a> 
         </span> 
         <span class='two' style='float:right; padding-right:5px;'> 
          <img class='left' width='24' height='24' src='img/posComment.png' /> Positive:18 
          <img src='img/negComment.png' width='24' height='24' /> Negative:10 
          <img width='24' height='24' src='img/neuComment.png' />Neutral:5 
         </span> 
        </div><!-- /professor_desc--> 
        <div id='clear'> </div> 
        <div class='professor_comments'> 
             "; // end echo 
       echo "     
       <h2 style='margin:0; padding:0;'>Recent Comments</h2> 
       <div class='comment'>by Anonymous on {$row2['date']}: <br> {$row2['info']} </div> 

        <input type='text' id='addComment' name='addComment' /> 
        "; 

       } else { 
        echo "No results."; 
       } 
       unset($sth); 
       ?> 
?> 

为什么没有这个拉动FNAME，LNAME等价值观？ ？

其拉动“By Anonoymous”部分，但没有任何值。任何人都

Answer 1

我不能对某些如果这些是你的问题的一部分，但我确实看到几件事情：

首先，您选择传递参数，当你调用execute绑定到的声明，但是没有占位符这样的参数 - 这应该产生大约绑定参数例外，我想：

$sth = $pdo->prepare(' 
      SELECT D.name, P.lname, P.fname, P.picpath, P.email, C.info, C.date 
       FROM Department D, Professor P, Comment C 
      WHERE D.dID = P.dID 
      AND P.pID = C.pID; 
     '); 
     $sth->execute(array(
      $pID // there is no placeholder to bind this too! 
     )); 

其次，除非你忽略周围的代码与你的循环，你不会循环通过你的结果正确，应该是这个样子：

if($sth->rowCount() > 0) { 
    while(false !== ($row = $sth->fetch(PDO::FETCH_ASSOC)) 
    { 
    // your output logic 
    } 
} 

最后您使用的是未定义的变量$row2 ...我只看到$row被分配在你的代码：

echo "     
       <h2 style='margin:0; padding:0;'>Recent Comments</h2> 
       <div class='comment'>by Anonymous on {$row2['date']}: <br> {$row2['info']} </div> 

        <input type='text' id='addComment' name='addComment' /> 
        ";